Investigate the different amounts of energy given off when different alcohols are combusted. I will be using the alcohols of methanol, ethanol, propanol, butanol and pentanol.

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Stuart Bird                                                               5/8/2007

TO FIND THE ENTHALPY OF THE COMBUSTION OF DIFFERENT ALCOHOLS

PLANNING

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

(Alcohol + Oxygen → Carbon Dioxide + Water)

Aim:

        I am going to investigate the different amounts of energy given off when different alcohols are combusted. I will be using the alcohols of methanol, ethanol, propanol, butanol and pentanol.

Safety note: In this experiment I will be dealing with Bunsen burners and the heating of alcohols. For this reason I wore safety goggles and I was very careful when handling the alcohols as they can be very dangerous.

Variables:

        Independent Variable: Type of alcohol

        Dependent Variable: Temperature rise of the water

        Fixed Variables: Volume of water (100cm3), length of time to heat water (2    

                                        mins),  length between the wick burner and the beaker of water(5cm).

Apparatus:

  1. Five different alcohol liquids (Methanol; Ethanol; Propanol; Butanol; Pentanol)
  2. Alcohol burners, contains with a wick down the centre so that the liquid can soak up. These were already set out by the technicians with the correct alcohol liquids before hand.
  3. Calorimeter (thermometer and a tin)I will be using a copper can for this experiment
  4. Stopwatch
  5. 100cm3 measuring cylinder
  6. Electric balance
  7. Heat mat and safety goggles

Diagram:

Method:

I set up the apparatus as show in the diagram

  • I measured out 100cm3 of water with a measuring cylinder and poured this into the tin and put the thermometer in too
  • I noted down the temperature of the water before I started the experiment and also the mass of the alcohol burner
  • I then set up the beaker in the boss head making sure that each alcohol burner was 5cm away from the beaker at all times.
  • I wanted to measure how much the temperature of the water rose over a period of 2 minutes so I heated the water for that amount of time
  • I then took the temperature after 2 minutes and also reweighed the alcohol burner
  • I did this experiment 5 times with methanol, ethanol, propanol, butanol and pentanol
  • I then did 2 repeats of all these alcohols in order to obtain more accurate results

 

Prediction:

   I predict that as the relative mass of the alcohol increases (i.e. a greater number of carbons) the enthalpy will increase.  Therefore Pentanol will release the greatest amount of energy and Methanol the least. I predict that the results will show a trend of proportionality- therefore the graph should show a straight line going through the origin.

I predict that the smaller carbons will burn with a smaller flame thus giving out less energy. Therefore the liquids with the more numbers of carbons for example Butanol and Pentanol will burn with a much larger flame thus giving out more energy.

   I can also work this prediction out using bond energies. I expect these bond energy results to show negative results, which means that the experiment is exothermic and is giving out heat. Although I know I must appreciate that my results will not be entirely accurate as heat will not completely be transferred to heating the water. Some will be lost to the surroundings etc.

All alcohols are bonded in the same way; a chain of carbons with an O-H bond on the end

  1. Methanol (CH3OH)        
  2. Ethanol (C2H5OH)
  3. Propanol (C3H7OH)
  4. Butanol (C4H9OH)
  5. Pentanol (C5H11OH)

Quantitative Prediction

Using bond energy diagrams I can work out the approximate ΔH for each alcohol.

Methanol                2CH3OH + 3O2 → 2CO2 + 4H2O

        Bond Breaking        Bond Making

        2 x        3(C-H)        =         6 x 412        2 x         2(C=O)        =        4 x 743

        2 x        (C-O)        =        2 x 360        4 x         2(O-H)        =        8 x 463

        2 x        (O-H)        =        2 x 463                        =        -6676

        3 x        (O=O)        =        3 x 496

                        =        5606

ΔH = -1070 Kj/mol

However as there are 2mols of Methanol

I must divide this number by 2– ΔH = -535 Kj/mol

Ethanol                        C2H5OH + 3O2 → 2CO2 + 3H2O

        Bond Breaking                Bond Making

                 5(C-H)        =        5 x 412        2 x        2(C=O)        =        4 x 743

                (C-C)        =        348        3 x        2(O-H)        =        6 x 463

                (C-O)        =        360                        =        -5750

                (O-H)        =        463

        3 x        (O=O)        =        3 x 496

                        =        4719

Join now!

                                                                    ΔH = -1031 Kj/mol

Propanol                        2C3H7OH + 9O2 → 6CO2 + 8H2O

        Bond Breaking                Bond Making

        2 x         7(C-H)        =        14 x 412        6 x        2(C=O)        =        12 x 743

        2 x         2(C-C)        =        4 x 348        8 x         2(O-H)        =        16 x 463

        2 x         (C-O)        =        2 x 360                        =         -16324

        2 x        (O-H)        =        2 x 463

        9 x        (O=O)        =        9 x 496

                        =        13270

ΔH = -3054/2

                                                  ...

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